Figure 1 shows a sketch of part of the curve with equation $y = 4x - xe^{-x}, \, x > 0$ The curve meets the x-axis at the origin O and cuts the x-axis at the point A - Edexcel - A-Level Maths Pure - Question 4 - 2015 - Paper 4

To evaluate the integral:

$\int xe^{-\frac{1}{2}x} \, dx$

Using integration by parts, we let:

$u = x, \, dv = e^{-\frac{1}{2}x} \, dx$

Then, $du = dx$ and $v = -2e^{-\frac{1}{2}x}$. The integration by parts gives:

$\int u \, dv = uv - \int v \, du$

Thus, we have:

$= -2xe^{-\frac{1}{2}x} - \int -2e^{-\frac{1}{2}x} \, dx$

Solving this, we will evaluate:

$\int e^{-\frac{1}{2}x} \, dx = -2e^{-\frac{1}{2}x} + C$

Combining terms, we get:

$-2xe^{-\frac{1}{2}x} + 4e^{-\frac{1}{2}x} + C.$

The area R can be computed using:

$A = \int_0^{a} (4x - xe^{-x}) \, dx$

where $a$ is the value found in part (a).\nIntegrating:

$A = \left[ 2x^2 - 2xe^{-x} + 2e^{-x} \right]_0^{a}$

Calculating bounds leads to:

$A = 2a^2 - 2ae^{-a} + 2e^{-a} - (0)$

Substituting $a = -2\ln(2)$ into the area expression provides the exact area in terms of ln2.